VECTOR CALCULUS
Prove the expression for∇·ain orthogonal curvilinear coordinates.Let us consider the sub-expression∇·(a 1 ˆe 1 ). Noweˆ 1 =eˆ 2 ׈e 3 =h 2 ∇u 2 ×h 3 ∇u 3. Therefore
∇·(a 1 ˆe 1 )=∇·(a 1 h 2 h 3 ∇u 2 ×∇u 3 ),
=∇(a 1 h 2 h 3 )·(∇u 2 ×∇u 3 )+a 1 h 2 h 3 ∇·(∇u 2 ×∇u 3 ).However,∇·(∇u 2 ×∇u 3 ) = 0, from (10.43), so we obtain
∇·(a 1 eˆ 1 )=∇(a 1 h 2 h 3 )·(
eˆ 2
h 2×
ˆe 3
h 3)
=∇(a 1 h 2 h 3 )·ˆe 1
h 2 h 3;
letting Φ =a 1 h 2 h 3 in (10.60) and substituting into the above equation, we find
∇·(a 1 ˆe 1 )=1
h 1 h 2 h 3∂
∂u 1(a 1 h 2 h 3 ).Repeating the analysis for∇·(a 2 ˆe 2 )and∇·(a 3 ˆe 3 ), and adding the results we obtain (10.61),
as required.
LaplacianIn the expression for the divergence (10.61), let
a=∇Φ=1
h 1∂Φ
∂u 1ˆe 1 +1
h 2∂Φ
∂u 2eˆ 2 +1
h 3∂Φ
∂u 3ˆe 3 ,where we have used (10.60). We then obtain
∇^2 Φ=1
h 1 h 2 h 3[
∂
∂u 1(
h 2 h 3
h 1∂Φ
∂u 1)
+∂
∂u 2(
h 3 h 1
h 2∂Φ
∂u 2)
+∂
∂u 3(
h 1 h 2
h 3∂Φ
∂u 3)]
,which is the expression for the Laplacian in orthogonal curvilinear coordinates.
CurlThe curl of a vector fielda=a 1 eˆ 1 +a 2 ˆe 2 +a 3 ˆe 3 in orthogonal curvilinear
coordinates is given by
∇×a=1
h 1 h 2 h 3∣
∣
∣
∣
∣
∣
∣
∣
∣h 1 eˆ 1 h 2 eˆ 2 h 3 ˆe 3∂
∂u 1∂
∂u 2∂
∂u 3
h 1 a 1 h 2 a 2 h 3 a 3∣
∣
∣
∣
∣
∣
∣
∣
∣. (10.62)
Prove the expression for∇×ain orthogonal curvilinear coordinates.Let us consider the sub-expression∇×(a 1 eˆ 1 ). Sinceˆe 1 =h 1 ∇u 1 we have
∇×(a 1 ˆe 1 )=∇×(a 1 h 1 ∇u 1 ),
=∇(a 1 h 1 )×∇u 1 + a 1 h 1 ∇×∇u 1.But∇×∇u 1 =0,soweobtain
∇×(a 1 eˆ 1 )=∇(a 1 h 1 )׈e 1
h 1